# Python – Append according to Kth character

Given a String list, append to String i or j value depending on Kth index value.

Input : test_list = [“geeksforgeeks”, “best”, “for”, “geeks”], K = 2, N = ‘e’, i, j = “@@”, “..”
Output : [‘geeksforgeeks..’, ‘best@@’, ‘for@@’, ‘geeks..’]
Explanation : geeksforgeeks and geeks having similar 2nd occ. value as ‘e’, hence gets appended by “..”.

Input : test_list = [“giiksforgeeks”, “bst”, “for”, “geeks”], K = 2, N = ‘e’, i, j = “@@”, “..”
Output : [‘giiksforgeeks@@’, ‘best@@’, ‘for@@’, ‘geeks@@’]
Explanation : No values with K value ‘e’, all appended by @@.

Method #1 : Using loop

This is brute way to solve this problem, we check for each string’s Kth index, if found to be N, then i value is appended else j is appended.

## Python3

 `# Python3 code to demonstrate working of  ` `# Append according to Kth character ` `# Using loop ` ` `  `# initializing lists ` `test_list ``=` `[``"geeksforgeeks"``, ``"best"``, ``"for"``, ``"geeks"``] ` ` `  `# printing string ` `print``(``"The original list : "` `+` `str``(test_list)) ` ` `  `# initializing K ` `K ``=` `2`  ` `  `# initializing N  ` `N ``=` `'e'` ` `  `# initializing i, j  ` `i, j ``=` `"**"``, ``"##"` ` `  `res ``=` `[] ` `for` `sub ``in` `test_list: ` `     `  `    ``# checking for Kth index to be N ` `    ``if` `sub[K] ``=``=` `N: ` `        ``res.append(sub ``+` `i) ` `    ``else` `: ` `        ``res.append(sub ``+` `j) ` ` `  `# printing results  ` `print``(``"The resultant List : "` `+` `str``(res)) `

Output

```The original list : ['geeksforgeeks', 'best', 'for', 'geeks']
The resultant List : ['geeksforgeeks**', 'best##', 'for##', 'geeks**']
```

Method #2 : Using list comprehension

This solves this problem in similar manner, just difference being, it’s a shorthand and can be used  as one liner approach to solve this problem.

## Python3

 `# Python3 code to demonstrate working of  ` `# Append according to Kth character ` `# Using list comprehension ` ` `  `# initializing lists ` `test_list ``=` `[``"geeksforgeeks"``, ``"best"``, ``"for"``, ``"geeks"``] ` ` `  `# printing string ` `print``(``"The original list : "` `+` `str``(test_list)) ` ` `  `# initializing K ` `K ``=` `2`  ` `  `# initializing N  ` `N ``=` `'e'` ` `  `# initializing i, j  ` `i, j ``=` `"**"``, ``"##"` ` `  `# shorthand to solve this problem ` `res ``=` `[sub ``+` `i ``if` `sub[K] ``=``=` `N ``else` `sub ``+` `j ``for` `sub ``in` `test_list] ` ` `  `# printing results  ` `print``(``"The resultant List : "` `+` `str``(res)) `

Output

```The original list : ['geeksforgeeks', 'best', 'for', 'geeks']
The resultant List : ['geeksforgeeks**', 'best##', 'for##', 'geeks**']
```

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